Question

The electric field in the region between two concentric charged spherical shells​

Answer 1

Explanation:

Let us find total charge enclosed in a sphere of radius r,

Q

′

=Q+∫

a

r

r

A

4πr

2

dr=Q+2πAr

2

−2πAa

2

By Gauss law,

E×4πr

2

=Q−2πAa

2

+2πAr

2

Given, E is independent of r

Hence, Q−2πAa

2

=0

This gives A=

2πa

2

Q

Answer 2

Answer:

Let us find total charge enclosed in a sphere of radius r,

Q  

′

=Q+∫  

a

r

​  

 

r

A

​  

4πr  

2

dr=Q+2πAr  

2

−2πAa  

2

 

By Gauss law,

E×4πr  

2

=Q−2πAa  

2

+2πAr  

2

 

Given, E is independent of r

Hence, Q−2πAa  

2

=0

This gives A=  

2πa  

2

 

Q

​

Explanation:

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